Solving a Diophantine equation involving fourth powers and squares

Question

I am trying to show that $x^4 - 3y^4=z^2$ has no solutions over the positive integers. I tried working on $\mathbb{Z}[\sqrt{3}]$ but couldn't go far. Factorizing didn't do much either.

Answer 1

Consider a positive integer solution, with $y$ minimal, of

$$ x^4-3y^4=z^2\tag{1}$$

We can suppose that $x,y,z$ are pairwise coprime since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation could occur. Then, considered modulo $8$ we see that $y$ is even. We rewrite the equation as

$$\left (\frac{x^2-z}{2}\right )\left (\frac{x^2+z}{2}\right)=12t^4.$$ Since the two bracketed factors, $L$ and $M$ say, differ by the odd integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $x$ and $z$, a contradiction.

Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=12$ and $t=uv$, with $u$ and $v$ coprime, and $$au^4+cv^4=x^2.$$ Since $x$ is odd the only possibility is $u^4+12v^4=x^2.$ (Note that $4u^4+3v^4=x^2$ is not possible modulo $4$). Then $$\left (\frac{u^2-x}{2}\right )\left (\frac{u^2+x}{2}\right)=-3v^4.$$ Since the two bracketed factors, $L$ and $M$ say, differ by the integer $x$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $x$ and $u$, a contradiction. Therefore $\{L,M\}=\{aU^4,cV^4\}$, where $ac=-3$ and $v=UV$, with $U$ and $V$ coprime, and $$aU^4+cV^4=u^2.$$ Modulo $3$ the only possibility is $\{a,c\}=\{1,-3\}.$ From a solution $(x,y,z)$ of equation (1) we have therefore deduced a second solution $(X,Y,Z)$, where $$Y=\frac{y}{2UX}<y,$$ a contradiction.

NOTE

This is the method used on

Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?